Electrostatic Potential and Capacitance - Result Question 45
47. If the potential of a capacitor having capacity $6 \mu F$ is increased from $10 V$ to $20 V$, then increase in its energy will be
(a) $4 \times 10^{-4} J$
(b) $4 \times 10^{-4} J$
(c) $9 \times 10^{-4} J$
(d) $12 \times 10^{-6} J$
[1995]
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Answer:
Correct Answer: 47. (c)
Solution:
- (c) Capacitance of capacitor $(C)=6 \mu F=6$ $\times 10^{-6} F$; Initial potential $(V_1)=10 V$ and final potential $(V_2)=20 V$.
The increase in energy $(\Delta U)$
$=\frac{1}{2} C(V_2^{2}-V_1^{2})$
$=\frac{1}{2} \times(6 \times 10^{-6}) \times[(20)^{2}-(10)^{2}]$
$=(3 \times 10^{-6}) \times 300=9 \times 10^{-4} J$.
